![]() ![]() This means that all of the points in the. Get the free view of Chapter 12, Reflection Concise Maths Class 10 ICSE additional questions for Mathematics Concise Maths Class 10 ICSE CISCE,Īnd you can use to keep it handy for your exam preparation. When we reflect a figure over the x-axis, we are essentially flipping the figure over a line parallel to the y-axis. Maximum CISCE Concise Maths Class 10 ICSE students prefer Selina Textbook Solutions to score more in exams. The questions involved in Selina Solutions are essential questions that can be asked in the final exam. Using Selina Concise Maths Class 10 ICSE solutions Reflection exercise by students is an easy way to prepare for the exams, as they involve solutionsĪrranged chapter-wise and also page-wise. ![]() Selina textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.Ĭoncepts covered in Concise Maths Class 10 ICSE chapter 12 Reflection are Reflection of a Point in a Line, Reflection of a Point in the Origin., Reflection Examples, Reflection Concept, Invariant Points. This will clear students' doubts about questions and improve their application skills while preparing for board exams.įurther, we at provide such solutions so students can prepare for written exams. Selina solutions for Mathematics Concise Maths Class 10 ICSE CISCE 12 (Reflection) include all questions with answers and detailed explanations. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. If P has co-ordinates (4, 6) evaluate a and b. A ray of light coming from the point (1,2) is reflected at a point A on the x-axis and then passes through the point (5,3). has the CISCE Mathematics Concise Maths Class 10 ICSE CISCE solutions in a manner that help students Below are three examples of reflections in coordinate plane. The point (a, b) is first reflected in the origin and then reflected in the y-axis to P. So the image (that is, point B) is the point (1/25, 232/25).Chapter 1: GST (Goods And Service Tax) Chapter 2: Banking (Recurring Deposit Account) Chapter 3: Shares and Dividend Chapter 4: Linear Inequations (In one variable) Chapter 5: Quadratic Equations Chapter 6: Solving (simple) Problems (Based on Quadratic Equations) Chapter 7: Ratio and Proportion (Including Properties and Uses) Chapter 8: Remainder and Factor Theorems Chapter 9: Matrices Chapter 10: Arithmetic Progression Chapter 11: Geometric Progression Chapter 12: Reflection Chapter 13: Section and Mid-Point Formula Chapter 14: Equation of a Line Chapter 15: Similarity (With Applications to Maps and Models) Chapter 16: Loci (Locus and Its Constructions) Chapter 17: Circles Chapter 18: Tangents and Intersecting Chords Chapter 19: Constructions (Circles) Chapter 20: Cylinder, Cone and Sphere Chapter 21: Trigonometrical Identities Chapter 22: Height and Distances Chapter 23: Graphical Representation Chapter 24: Measure of Central Tendency(Mean, Median, Quartiles and Mode) Chapter 25: Probability So the intersection of the two lines is the point C(51/50, 457/50). The final figure will be an equal distance. A reflection reverses the object’s orientation relative to the given line. The most frequently used lines are the y-axis, the x-axis, and the line y x, though any straight line will technically work. Using the substitution method gives 7x + 2 = (-1/7)x + 65/7 (50/7)x = 51/7 x = 51/50. A reflection in geometry is a mirror image of a function or object over a given line in the plane. Now we need to find the intersection of the lines y = 7x + 2 and y = (-1/7)x + 65/7 by solving this system of equations. So the equation of this line is y = (-1/7)x + 65/7. So the desired line has an equation of the form y = (-1/7)x + b. Since the line y = 7x + 2 has slope 7, the desired line (that is, line AB) has slope -1/7 as well as passing through (2,9). ![]() So we first find the equation of the line through (2,9) that is perpendicular to the line y = 7x + 2. ![]() Then, using the fact that C is the midpoint of segment AB, we can finally determine point B.Įxample: suppose we want to reflect the point A(2,9) about the line k with equation y = 7x + 2. Then we can algebraically find point C, which is the intersection of these two lines. So we can first find the equation of the line through point A that is perpendicular to line k. Note that line AB must be perpendicular to line k, and C must be the midpoint of segment AB (from the definition of a reflection). Let A be the point to be reflected, let k be the line about which the point is reflected, let B represent the desired point (image), and let C represent the intersection of line k and line AB. ![]()
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